Giai pt:\(\sqrt{x-2}+\sqrt{y+2003}+\sqrt{z-2004}=\frac{1}{2}\left(x+y+z\right)\)
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\(\frac{\sqrt{x-2002}}{x-2002}-\frac{1}{x-2002}+\frac{\sqrt{y-2003}}{y-2003}-\frac{1}{y-2003}+\frac{\sqrt{z-2004}}{z-2004}-\frac{1}{z-2004}=\frac{3}{4}\)
\(1-\frac{1}{x-2002}+1-\frac{1}{y-2003}+1-\frac{1}{z-2004}=\frac{3}{4}\)
\(3-\frac{1}{x-2002}-\frac{1}{y-2003}-\frac{1}{z-2004}=\frac{3}{4}\)
\(\frac{1}{x-2002}+\frac{1}{y-2003}+\frac{1}{z-2004}=3-\frac{3}{4}=\frac{9}{4}\)
=> không có giá trị x,y,z thỏa mãn đề
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Bạn tham khảo tại đây:
Câu hỏi của Vũ Sơn Tùng - Toán lớp 9 | Học trực tuyến
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Phương trình có vô số nghiệm
Nếu thay \(\sqrt{y-2008}\) bằng \(\sqrt{y+2008}\) thì phương trình có bộ nghiệm duy nhất: \(\left(x;y;z\right)=\left(2010;-2007;3\right)\)
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+ \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=4\Rightarrow x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
+ \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)\)
+ Tương tự : \(y+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\); \(z+1=\left(\sqrt{x}+\sqrt{z}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+ \(P=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}+\sqrt{z}\right)^2\left(\sqrt{z}+\sqrt{x}\right)^2}\cdot\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=2\)
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\(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)
\(\Leftrightarrow x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}=4\)
\(\Leftrightarrow2+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
Khi đó ta có : \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow x+1=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(\Leftrightarrow x+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
Tương tự : \(y+1=\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\);
\(z+1=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Ta lần lượt xét các biểu thức :
+) \(\sqrt{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\sqrt{\left[\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\right]^2}\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+) \(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
Do đó ta có :
\(P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\cdot\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(P=2\)
Vậy...
\(\Leftrightarrow x+y+z=2\sqrt{x-2}+2\sqrt{y+2003}+2\sqrt{z-2004}\)
\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2003-2\sqrt{y+2003}+1\right)\)
\(+\left(z-2004-2\sqrt{z-2004}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2003}-1\right)^2+\left(\sqrt{z-2004}-1\right)^2=0\)
Vì biểu thức trên là tổng của các số hạng không âm nên nó bằng 0 khi và chỉ khi các số hạng phải bằng 0
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-2003}=1\\\sqrt{z-2004}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2004\\z=2005\end{cases}}}\)
\(ĐK:x\ge2,y\ge-2003,z\ge2004\)
Pt đã cho tương đương :
\(x+y+z-2\sqrt{x-2}-2\sqrt{y+2003}-2\sqrt{z-2004}=0\)
\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2003-2\sqrt{y+2003}+1\right)+\left(z-2004-2\sqrt{z-2004}+1\right)\)\(=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2003}-1\right)^2+\left(\sqrt{z-2004}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=1\\y+2003=1\\z-2004=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=-2002\\z=2005\end{cases}}\)(Thỏa mãn)